Spacecraft and Aircraft Dynamics

1 Spacecraft and Aircraft DynamicsMatthew M. PeetIllinois Institute of TechnologyLecture 9: 6 DOF Equations of MotionAircraft DynamicsLecture 9In this Lecture we will cover:Newton s Laws ~Mi=ddt~H ~Fi=mddt~vRotating Frames of Reference Equations of Motion in body -Fixed Frame Often ConfusingM. PeetLecture 9:2 / 24 Review: Coordinate RotationsPositive DirectionsIf in doubt, use the right-hand :Positive DirectionsFigure:Positive RotationsM. PeetLecture 9:3 / 24 Review: Coordinate RotationsRoll-Pitch-YawThere are 3 basic rotations an Aircraft can make: Roll = Rotation aboutx-axis Pitch = Rotation abouty-axis Yaw = Rotation aboutz-axis Each rotation is a one-dimensional two coordinate systems can be related by a sequence of 3 PeetLecture 9:4 / 24 Review: Forces and MomentsForcesThese forces and moments have standard labels.

2 The Forces are:XAxial ForceNet Force in the positivex-directionYSide ForceNet Force in the positivey-directionZNormal Force Net Force in the positivez-directionM. PeetLecture 9:5 / 24 Review: Forces and MomentsMomentsThe Moments are called, intuitively:LRolling Moment Net Moment in the positivep-directionMPitching Moment Net Moment in the positiveq-directionNYawing Moment Net Moment in the positiver-directionM. PeetLecture 9:6 / 246 DOF: Newton s LawsForcesNewton s Second Law tells us that for a particleF=ma. In vector form:~F= i~Fi=mddt~VThat is, if~F= [FxFyFz]and~V= [u v w], thenFx=mdudtFx=mdvdtFz=mdwdtDefinition 1.~L=m~Vis referred to asLinear s Second Law is only valid if~Fand~Vare defined in anInertialcoordinate coordinate system isInertialif it is not accelerating or PeetLecture 9:7 / 24 Newton s LawsMomentsUsing Calculus, this concept can be extended to rigid bodiesby integration overall particles.

3 ~M= i~Mi=ddt~HDefinition ~H= (~rc ~vc)dmis theangular momentum of a rigid body can be found as~H=I~ Iwhere~ I= [p, q, r]Tis the angular rotation vector of the body about thecenter of mass. pis rotation about thex-axis. qis rotation about they-axis. ris rotation about thez-axis. Iis defined in matrixIis theMoment of Inertia PeetLecture 9:8 / 24 Newton s LawsMoment of InertiaThe moment of inertia matrix is defined asI= Ixx Ixy Ixz IyxIyy Iyz Izx IzyIzz Ixy=Iyx= xydmIxx= (y2+z2)dmIxz=Izx= xzdmIyy= (x2+z2)dmIyz=Izy= yzdmIzz= (x2+y2)dmSo HxHyHz = Ixx Ixy Ixz IyxIyy Iyz Izx IzyIzz pIqIrI wherepI,qIandrIare the rotation vectors as expressed in the inertial framecorresponding PeetLecture 9:9 / 24 Moment of InertiaExamples:Homogeneous SphereIsphere=25mr2 1 0 00 1 00 0 1 RingIring=mr2 120 001200 0 1 M.

4 PeetLecture 9:10 / 24 Moment of InertiaExamples:Homogeneous DiskIdisk=14mr2 1 +13hr20001 +13hr200012 F/A-18I= 0 kslug ft2M. PeetLecture 9:11 / 24 Problem:The body -Fixed FrameThe moment of inertia matrix,I, is fixed in the body -fixed frame. However,Newton s law only applies for an inertial frame:~M= i~Mi=ddt~HIf the body -fixed frame is rotating with rotation vector~ , then for any vector,~a,ddt~ain the inertial frame isd~adt I=d~adt B+~ ~aSpecifically, for Newton s Second Law~F=md~Vdt B+m~ ~Vand~M=d~Hdt B+~ ~HM. PeetLecture 9:12 / 24 Equations of MotionThus we have FxFyFz =m u v w +mdet x y zp q ru v w =m u+qw rv v+ru pw w+pv qu and LMN = Ixx Ixy Ixz IyxIyy Iyz Izx IzyIzz p q r +~ Ixx Ixy Ixz IyxIyy Iyz Izx IzyIzz pqr = Ixx p Ixy q Ixz r Ixy p+Iyy q Iyz r Ixz p Iyz q+Izz r +~ pIxx qIxy rIxz pIxy+qIyy rIyz pIxz qIyz+rIzz = Ixx p Ixy q Ixz r+q( pIxz qIyz+rIzz) r( pIxy+qIyy rIyz) Ixy p+Iyy q Iyz r p( pIxz qIyz+rIzz) +r(pIxx qIxy rIxz) Ixz p Iyz q+Izz r+p( pIxy+qIyy rIyz) q(pIxx qIxy rIxz) Which is too much for any mortal.

5 For Aircraft , we havesymmetryabout thex-z plane. ThusIyz=Ixy= 0. Spacecraft ?M. PeetLecture 9:13 / 24 Equations of MotionReduced EquationsWithIxy=Iyz= 0, we have, in summary: FxFyFz =m u+qw rv v+ru pw w+pv qu and LMN = Ixx p Ixz r qpIxz+qrIzz rqIyyIyy q+p2 Ixz prIzz+rpIxx r2 Ixz Ixz p+Izz r+pqIyy qpIxx+qrIxz Right now, Translational variables (u,v,w) depend on rotational variables (p,q,r). Rotational variables (p,q,r) do not depend on translational variables(u,v,w).IFor Aircraft , however, Moment forces (L,M,N) depend on rotational andtranslational PeetLecture 9:14 / 24 EOMs in Rotating FrameExample: SnipersQuestion:Consider a sniper firing a rifle due east at the equator. Ignoringgravity and drag, what are the equations of motion of the bullet?

6 Use theNorth-East-Up local coordinate system. Muzzle velocity:1000m/s. :The earth is rotating about its axis at angular velocity2 radday, The rotation is positive about the local North-axis. Thus~ = pqr = .000072700 Since the bullet is in free-flight, there are no forces. Thus the Equations ofmotion are 000 =m u+qw rv v+ru pw w+pv qu =m u v pw w+pv M. PeetLecture 9:15 / 24 EOMs in Rotating FrameExample: SnipersSimplified EOMs:Usingq=r= 0, we simplify to u= 0 v=pw w= :For initial conditionu(0) = 0,v(0) =Vandw(0) = 0has solutionu(t) = 0v(t) =v(0) cos(pt)w(t) = v(0) sin(pt)Sincepis very small compared to flight time, we can approximateu(t) = 0v(t) =v(0)w(t) = v(0)ptWhich yields displacementN(t) = 0E(t) =v(0)t U(t) = 12v(0)pt2 Conclusion:For a target at rangeE(ti) = 4km, we haveti= 4sand hence theerror at target is:N(ti) = 0U(ti) = 12 2000.

7 0000727 16 = course, if we were firing west, the error would be+ PeetLecture 9:16 / 24 Euler AnglesIssue:Equations of motion are expressed in the body -Fixed :How do determine rotation and velocity in the inertial frame. Forintercept, obstacle avoidance, :From Lecture 4, any two coordinate systems can be related througha sequence of three rotations. Recall these transformations are:Roll Rotation ( ) :R1( )= 1 000 cos sin 0 sin cos Pitch Rotation ( ):R2( )= cos 0 sin 01 0 sin 0 cos Yaw Rotation ( ):R3( )= cos sin 0sin cos 0001 M. PeetLecture 9:17 / 24 Euler AnglesDefinition termEuler Anglesrefers to the angles of rotation ( , , ) needed to gofrom one coordinate system to another using the specific sequence of rotationsYaw-Pitch-Roll:~VBF=R1( )R2( )R3( )~ BENE:Euler angles are often defined differently ( 3-1-3).

8 We usethe book composite rotation matrix can be writtenR1( )R2( )R3( ) = 1 000 cos sin 0 sin cos cos 0 sin 01 0 sin 0 cos cos sin 0sin cos 0001 This moves a vectorInertial Frame body -Fixed FrameM. PeetLecture 9:18 / 24 Euler AnglesTo move a vectorBody-Fixed Frame Inertial Framewe need toInvert the Rotations. Rotation matrices are easily inverted, howeverRi( ) 1=Ri( )Thus~VI= (R1( )R2( )R3( )) 1~VBF, where(R1( )R2( )R3( )) 1=R3( ) 1R2( ) 1R1( ) 1=R3( )R2( )R1( )= cos sin 0sin cos 0001 cos 0 sin 01 0 sin 0 cos 1 000 cos sin 0 sin cos = cos cos sin sin cos cos sin cos sin cos + sin sin cos sin sin sin sin + cos cos cos sin sin sin cos sin sin cos cos cos These transformations now describe PeetLecture 9:19 / 24 Euler AnglesVelocity vectorThus to find the inertial velocity vector, we must rotateFROMthe body -fixedcoordinates to the inertial frame: dxdtdydtdzdt =R3( )R2( )R1( ) uvw M.

9 PeetLecture 9:20 / 24 Euler AnglesThe rate of rotation of the Euler Angles can be found by rotating the rotationvector into the inertial frame pqr = 10 sin 0 cos cos sin 0 sin cos cos This transformation can also be reversed as = 1 sin tan cos tan 0cos sin 0 sin sec cos sec pqr M. PeetLecture 9:21 / 24 SummaryM. PeetLecture 9:22 / 24 ConclusionIn this lecture we have coveredEquations of Motion How to differentiate Vectors in Rotating Frames Derivation of the Nonlinear 6 DOF Equations of MotionEuler Angles Definition of Euler Angles Using Rotation Matrices to transform vectors Derivatives of the Euler anglesIRelationship top-q-rin body -Fixed FrameM. PeetLecture 9:23 / 24 Next LectureIn the next lecture we will coverLinearized Equations of Motion How to linearize the nonlinear 6 DOF EOM How to linearize the force and moment contributionsForce and Moment Contributions The gravity and thrust contributions The full linearized equations of motion including forces and moments How to decouple into Longitudinal and Lateral DynamicsIReminder on how to create a state-space PeetLecture 9:24 / 24